\(t\sin(2t)\)的拉普拉斯变换为 A: \(\dfrac{2}{s^2+4}\) B: \(\dfrac{s}{s^2+4}\) C: \(\dfrac{4}{(s^2+4)^2}\) D: \(\dfrac{4s}{(s^2+4)^2}\)

如果简单正向闭曲线L所围成区域的面积为S，那么$S = (\quad ).$ A: $\dfrac{1}{2}\oint_L {xdx - ydy} $ B: $\dfrac{1}{2}\oint_L {ydy - xdx} $ C: $\dfrac{1}{2}\oint_L {ydx - xdy} $ D: $\dfrac{1}{2}\oint_L {xdy - ydx} $

一粒子在一维无限深方势阱中运动而处于基态。试求(1) 坐标$x$的平均值 A: $a$ B: $\dfrac{a}{2}$ C: $\dfrac{a}{4}$ D: $\dfrac{3a}{4}$

（接上题）（2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$

（2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$

采样系统的闭环传递函数为\(G_{\rm CL}(z)=2z^{-1}-z^{-2}\)，则该系统的单位阶跃响应为 A: \(\dfrac{z-1}{z^2-z}\) B: \(\dfrac{2z-1}{z^2-z}\) C: \(\dfrac{2z-1}{z^2-1}\)

$n=1$时，经典图像中电子围绕质子作半径为$a_0$的圆周运动时的总能量。(1) 总能量为： A: $E=\dfrac{e^2}{8\pi \epsilon_0 a_0}$ B: $E=-\dfrac{e^2}{4\pi \epsilon_0 a_0}$ C: $E=\dfrac{e^2}{4\pi \epsilon_0 a_0}$ D: $E=-\dfrac{e^2}{8\pi \epsilon_0 a_0}$

优学院: 曲面(xcosz+ycosx-dfrac {pi} {2}z=dfrac {pi} {2})在点((dfrac {pi} {2},1-dfrac {pi} {2},0))处的切平面方程是（ ）

下列函数在给定区间上满足罗尔定理条件的是( ). A: $f(x)=\dfrac 1{x},\; [-2,0]$ B: $f(x)=(x-4)^2,\;[-2,4]$ C: $f(x)=\sin x,\; [-\dfrac{3\pi}{2},\dfrac{\pi}{2}]$ D: $f(x)=|x|,\; [-1,1]$

已知\(L[e^{-at}]=\dfrac{1}{s+a}\)，则\(L[te^{-at}]\)为（ ）